Answer:
dV =3.39x10^-38 m/s
Explanation:
Consider a 2250-lb automobile clocked by law-enforcement radar at a speed of 85.5 mph (miles/hour). if the position of the car is known to within 5.0 feet at the time of the measurement, what is the uncertainty in the velocity of the car?
According to the Heisenberg uncertainty law which states tat te
product of uncertainty in position and uncertainty in momentum will be constant
\Delta x . \Delta P = \frac{h}{4\pi}
\Delta x . m \Delta v = \frac{h}{4\pi}
Mass(automobile)= 2150 lbs
1lbs=0.453592kg so:
2250 lbs= 1020.3 kg
Uncertainty(delta X)= 5ft
1ft= 0.3048m so:
5ft= 1.524m
h= Planck's constant = 6.62606957 × 10-34 m2 kg / s
(delta V) = (delta P)/ Mass
Solving the original equation we get:
(delta V) = h/( (4pi)(Mass)(delta X) )
= (6.626X10^-34)/( (4pi)( 1020.3 kg )(1.524m)
dV =3.39x10^-38 m/s
the uncertainty in the velocity of the car is dV =3.39x10^-38 m/s
The uncertainty in the velocity of the car during the measurement is 3.4 x 10⁻³⁸ m/s.
The given parameters;
The uncertainty in the velocity of the automobile is determined using Heisenberg uncertainty principle;
[tex]\Delta P \Delta x \geq \frac{h}{4\pi} \\\\m\Delta v\Delta x \geq \frac{h}{4\pi}\\\\\Delta v \geq \frac{h}{4\pi \times m \Delta x}\\\\\Delta v \geq \frac{6.626 \times 10^{-34}}{4\pi \times (1020.58)\times (1.52) }\\\\\Delta v \geq 3.4 \times 10^{-38} \ m/s[/tex]
Thus, the uncertainty in the velocity of the car during the measurement is 3.4 x 10⁻³⁸ m/s.
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